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Math Help - Derivatives using ln

  1. #1
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    Derivatives using ln

    Got a question here....

    Find the derivative of :

    P(t) = 500 / (1 + e^-t)

    I started by re-writing it in the form :

    P(t) = 500 (1 + e^-t)^-1

    Using the chain rule I came up with :

    P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'
    = -500(1 + e^-t)^-2(e^-t)(ln e)
    = -500(1 + e^-t)^-2(e^-t)(1)

    The question goes on to ask for the derivative at P'(3)

    P'(3) = -500(1 + e^-3)^-2(e^-3)
    = -500(.9074)(.04979)
    = -22.6

    I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.
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  2. #2
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    Quote Originally Posted by Jools View Post
    Got a question here....

    Find the derivative of :

    P(t) = 500 / (1 + e^-t)

    I started by re-writing it in the form :

    P(t) = 500 (1 + e^-t)^-1

    Using the chain rule I came up with :

    P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'
    = -500(1 + e^-t)^-2(e^-t)(ln e)


    From where did you bring that \ln e=1 ? It can only confuse you.

    = -500(1 + e^-t)^-2(e^-t)(1)

    The question goes on to ask for the derivative at P'(3)

    P'(3) = -500(1 + e^-3)^-2(e^-3)


    Here's the mistake: you actually have P'(3)=-500(1+e^{-3})(-2e^{-3})=1,000(1+e^{-3})e^{-3} , which is positive indeed.

    Tonio


    In fact now I realized that that -2 is a power (!!), but still: you wrote (1+e^{-t})'=e^{-t} , which is lacking a minus sign there!

    Tonio



    = -500(.9074)(.04979)
    = -22.6

    I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.
    .
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  3. #3
    Newbie Riyzar's Avatar
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    Quote Originally Posted by Jools View Post
    Got a question here....

    Find the derivative of :

    P(t) = 500 / (1 + e^-t)

    I started by re-writing it in the form :

    P(t) = 500 (1 + e^-t)^-1

    Using the chain rule I came up with :

    P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'
    = -500(1 + e^-t)^-2(e^-t)(ln e)
    = -500(1 + e^-t)^-2(e^-t)(1)

    The question goes on to ask for the derivative at P'(3)

    P'(3) = -500(1 + e^-3)^-2(e^-3)
    = -500(.9074)(.04979)
    = -22.6

    I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.
    (1+e^{-t})' does not equal (e^{-t})ln e
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Jools View Post
    Got a question here....

    Find the derivative of :

    P(t) = 500 / (1 + e^-t)

    I started by re-writing it in the form :

    P(t) = 500 (1 + e^-t)^-1

    Using the chain rule I came up with :

    P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'

    500(-1)(1+e^-t)^-2 (-(e^-t)) = 500 (1+e^-t)^-2 (e^-t)

    The question goes on to ask for the derivative at P'(3)

    P'(3) = -500(1 + e^-3)^-2(e^-3)
    = -500(.9074)(.04979)
    = -22.6
    Since the derivative is 500(1 + e^-3)^-2(e^-3), your value should not come up negative now
    I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.
    The derivative of \frac{500}{1 + e^{-t}} is \frac {500{e^{-t}}}{(1+e^{-t})^2}

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  5. #5
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    Would I also be correct in writing it like this:

    -500(1 + e^-t)^-2 (-t e^-t-1)

    Then solving for P '(3)?

    It doesn't give the same answer when I solve for P '(3) as I get with your method, so I am assuming no...
    Last edited by Jools; April 7th 2010 at 04:23 PM.
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